JEE MAIN - Physics (2021 - 24th February Morning Shift - No. 20)
A ball with a speed of 9 m/s collides with another identical ball at rest. After the collision, the direction of each ball makes an angle of 30$$^\circ$$ with the original direction. The ratio of velocities of the balls after collision is x : y, where x is __________.
Answer
1
Explanation
The situation is shown below
_24th_February_Morning_Shift_en_20_1.png)
Using conservation of linear momentum in y-direction,
pi = pf
As, pi = 0
and pf = mv1 sin30$$^\circ$$ $$-$$ mv2 sin30$$^\circ$$
$$\Rightarrow$$ 0 = m $$\times$$ $${1 \over 2}$$v1 $$-$$ m $$\times$$ $${1 \over 2}$$v2
$$\Rightarrow$$ v1 = v2 or v1 : v2 = 1 : 1
Since, v1 : v2 = x : y (given)
$$\therefore$$ x = 1
Using conservation of linear momentum in y-direction,
pi = pf
As, pi = 0
and pf = mv1 sin30$$^\circ$$ $$-$$ mv2 sin30$$^\circ$$
$$\Rightarrow$$ 0 = m $$\times$$ $${1 \over 2}$$v1 $$-$$ m $$\times$$ $${1 \over 2}$$v2
$$\Rightarrow$$ v1 = v2 or v1 : v2 = 1 : 1
Since, v1 : v2 = x : y (given)
$$\therefore$$ x = 1
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